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College Algebra Worksheets By O. Pauline Chow HACC Math 103 College Algebra Lecture Notes on Chapter 5.1-5.2 (audio dated 9/28/07 at 6:40 pm, 32 minutes long) Graphing method Substitution method Addition method Applications 3x3 systems Systems of Linear Equations In this chapter, you will review the methods of solving 2x2 and 3x3 systems of linear equations. For 2x2 systems, you will solve using graphing, substitution, and addition method. For 3x3 systems, you use the addition method. Graphing Method Ex 1: Solve the 2x2 system using graphing method. y = –2x + 6 4x + 2y = 4 For each equation, you need to find at least two points to graph. The graphs of the two linear equations are two parallel lines. Since they do not intersect, there is no solution to the system and we say the system is inconsistent. Ex 2: Solve the 2x2 system using graphing method. 2x – y = 4 x + 2y = –3 Note: When two lines intersect at a point, the point is called the point of intersection and we say the system is consistent. When two lines overlap each other, every point on the two lines is a point of intersection. We say the system is dependent and there is an infinite number of solutions. College Algebra Worksheets By O. Pauline Chow HACC Substitution Method Ex 1: Solve the 2x2 system using substitution method. –2x + 5y = 14 7x + 6y = –2 (1) (2) You need to solve for one of the two variables using one of the two equations, say, y in the first equation. –2x + 5y = 14 5y = 2x + 14 Add 2x to each side. y = 2 x + 14 (3) 5 Now substitute equation (3) into equation (2). 7x + 6 (2x + 14) = –2 5 35x + 6(2x + 14) = –10 35x + 12x + 84 = –10 47x + 84 = –10 47x = –94 x = –2 Multiply each side by 5. Combine like terms on the left side. Subtract 84 from each side. Divide each side by 47 Now substitute –2 for x into equation (3). y = 2(–2) + 14 = 10 = 2 5 5 So, the solution to the system is (–2, 2). The system is consistent. Ex 2: Solve the 2x2 system using substitution method. x + 3y = –3 5x – 6y = 27 (1) (2) Solve equation (1) for x. x = –3 – 3y (3) Substitute equation (3) into (2). 5(–3 – 3y) – 6y = 27 –15 – 15y – 6y = 27 –21y = 42 y = –2 Substitute –2 for y into equation (3) x = –3 – 3(–2) = –3 + 6 = 3 The solution to the system is (3, –2). Addition Method Ex 1: Solve the 2x2 system using addition method. –2x + 5y = 14 7x + 6y = –2 (1) (2) College Algebra Worksheets By O. Pauline Chow HACC You need to multiply an appropriate number to each equation, so that the coefficients of one of the two variables will add up to zero. So, you multiply 7 to equation (1) and 2 to equation (2). 7(–2x + 5y = 14) 2(7x + 6y = –2) –14x + 35y = 98 14x + 12y = –4 Now you add the two equations. –14x + 35y = 98 + 14x + 12y = –4 47y = 94 → y = 2 Now substitute 2 for y in one of the two equations, say (1) to solve for x. –2x + 5(2) = 14 –2x + 10 = 14 –2x = 4 x = –2 So, the solution to the system is (–2, 2). Ex 2: Solve the 2x2 system using addition method. –4x + 3y = –31 3x – 2y = 22 (1) (2) We will multiply 3 to equation (1) and 4 to equation (2) and add. 3(–4x + 3y = –31) 4(3x – 2y = 22) –12x + 9y = –93 12x – 8y = 88 y = –5 Now substitute –5 for y into equation (2) to solve for x. 3x – 2(–5) = 22 3x + 10 = 22 3x = 12, x = 4 The solution to the system is (4, –5). Ex 3: Solve the 2x2 system using addition method. 2x – y = –3 6x – 3y = 1 (1) (2) We will multiply –3 to equation (1) and add to equation (2). –3(2x – y = –3) 6x – 3y = 1 –6x + 3y = 9 6x – 3y = 1 0 = 10, impossible College Algebra Worksheets By O. Pauline Chow HACC There is no solution. The system is inconsistent. Applications Ex 1: Tanya got 6 paperbacks and 3 hardbacks for $8.25. Gretta got 4 paperbacks and 5 hardbacks for $9.25. Find the price for each. What is the total price for 7 paperbacks and 9 hardbacks? Let the price of one paperback be $x and that of hardback be $y. Since Tanya got 6 paperbacks and 3 hardbacks for $8.25, the first equation is 6x + 3y = 8.25 Since Gretta got 4 paperbacks and 5 hardback for $9.25, the second equation is 4x + 5y = 9.25 6x + 3y = 8.25 4x + 5y = 9.25 (1) (2) Multiply –2 to equation (1) and add to 3 times equation (2). –12x – 6y = –16.5 12x + 15y = 27.75 9y = 11.25 y = 1.25, price of one hardback is $1.25 6x + 3(1.25) = 8.25 6x + 3.75 = 8.25 6x = 4.5 x = 0.75, price of one paperback is $0.75 Cost of 7 paperbacks and 9 hardbacks is 7(0.75) + 9(1.25) or $16.5. Ex 2: In chemistry lab, Stephanie has to make a 35 milliliter solution that is 12% HCL. All that is in the lab is 8% and 15% HCL. How many milliliters of each solution should she use to obtain the desired mix? % of HCL Volume of solution Volume of HCL 8% x ml 0.08x ml 15% y ml 0.15y ml 12% 35 ml 0.12(35) ml x + y = 35 x = 35 – y 0.08x + 0.15y = 0.12(35) 8x + 15y = 420 Substitute equation (1) into equation (2). 8(35 – y) + 15y = 420 280 – 8y + 15y = 420 7y = 140, y = 20 (1) (2) x = 35 – 20 = 15 Stephanie should mix 15 milliliters of 8% HCL and 20 milliliters of 15% HCL. Ex 3 A private pilot flies a Cessna 172 on a trip that is 480 miles each way. It takes her 3 hours to get there with tailwind and 4 hours to return with headwind. What are the speed of the Cessna in still air and the speed of the wind? First you let x mph be the speed of the Cessna in still air and y mph be the speed of the wind. tail wind head wind distance 480 miles 480 miles speed (x + y) mph (x – y) mph time 3 hours 4 hours College Algebra Worksheets By O. Pauline Chow HACC Now apply the formula: distance = time * speed 3(x + y) = 480 x + y = 160 4(x – y) = 480 x – y = 120 x + y = 160 x – y = 120 2x = 280, x = 140 y = 160 – 140 = 20 Speed of the Cessna is 140 mph in still air and speed of the wind is 20 mph. 3x3 System To solve a 3x3 system, you need to use the addition method repeatedly to eliminate one of the 3 variables in the system and reduce the 3x3 system to a 2x2 system. Ex 1: Solve a 3x3 system. 2x – 6y + 4z = 8 3x – 9y + 6z = 12 5x – 15y + 10z = 20 (1) (2) (3) In order to keep track of the steps, starting with the first variable, say x. Use equations (1) and (2) to eliminate x by multiplying equation (1) by 3 and add to (–2) times equation (2). Use equations (1) and (3) to eliminate x by multiplying equation (1) by 5 and add to (–2) times equation (3). 3* eqt(1) + (–2)*eqt(2): 5*eqt(1) + (–2)*eqt(3): 6x – 18y + 12z = 24 –6x + 18y – 12z = –24 0=0 10x – 30y + 20z = 40 –10x + 20y – 20y = –40 0=0 Since both give an identity, the system is dependent, i.e. all three equations are equivalent to equation (1). The solution set is {2x – 6y + 4z = 8, where x, y and z are any real numbers} Ex 2: –2x + y – 3z = 6 4x – y + z = 2 2x – y + 3z = 1 (1) (2) (3) 2* (1) + (2): –4x + 2y – 6z = 12 + 4x – y + z = 2 y – 5z = 14 (1) + (3): –2x + y – 3z = 6 + 2x – y + 3z = 1 0 = 7 impossible The system is inconsistent and there is no solution. Ex 3: 2x + 3y + 5z = –14 3x – 2y + z = 5 –5x + y – 3z = –2 –3* eqt(1) + (2)*eqt(2): –6x – 9y – 15z = 42 + 6x – 4y + 2z = 10 –13y – 13z = 52 y + z = -4 (4) (1) (2) (3) 5*eqt(1) + (2)*eqt(3): 10x + 15y + 25z = –70 + –10x + 2y – 6z = –4 17y + 19z = –74 (5) College Algebra Worksheets By O. Pauline Chow HACC –17* eqt(4) + eqt(5) –17y – 17z = 68 17y + 19z = –74 2z = –6, z = –3 Substitute –3 for z into equation (4). y + (–3) = -4 y = –1 Substitute –3 for z and –1 for y into equation (1) to solve for x. 2x + 3y + 5z = –14 2x + 3(–1) + 5(–3) = –14 2x – 3 – 15 = –14 2x = 4, x = 2 The solution to the system is (2, –1, –3). Ex 4: 100x + 200y + 500z = 47 350x + 5y + 250z = 33.9 200x + 80y + 100z = 23.4 (1) (2) (3) 7* eqt(1) + (–2)*eqt(2): 700x + 1400y + 3500z = 329 + –700x – 10y – 500z = –67.8 1390y + 3000z = 261.2 (4) –2*eqt(1) + eqt(3): –200x – 400y – 1000z = –94 + 200x + 80y + 100z = 23.4 –320y – 900z = –70.6 Next you will need to eliminate y. 32*eqt(4) + 139*eqt(5): 44480y + 96000z = 8358.4 + –44480y – 125100z = –9813.4 –29100z = –1455 z = 0.05 Now substitute 0.05 for z in either (4) or (5) to solve for y. 1390y + 3000(0.05) = 261.2 1390y + 150 = 261.2 1390y = 111.2 y = 0.08 Substitute both values of y and z into (1), (2), or (3) to solve for x. 100x + 200(0.08) + 500(0.05) = 47 100x + 16 + 25 = 47 100x = 6 x = 0.06 The solution to the system is (0.06, 0.08, 0.05). Be sure to check the solution using (2) and (3). (5)